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# 数学代写|随机过程Stochastic Porcesses代考|AMATH562 Queues with many servers

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## 数学代写|随机过程Stochastic Porcesses代考|The model M/M/s

An important generalization of the $M / M / 1$ model is obtained by supposing that there are $s$ servers in the system and that they all serve at an exponential rate $\mu$. The other basic assumptions that were made in the description of the $M / M / 1$ model remain valid. Thus, the customers arrive in the system according to a Poisson process with rate $\lambda$. The capacity of the system is infinite, and the service policy is that by default, namely, first-come, firstserved.

We suppose that the arriving customers form a single queue and that the customer at the front of the queue advances to the first server who becomes available. A system with this waiting discipline is clearly more efficient than one in which there is a queue in front of each server, since, in this case, there could be one or more idle servers while some customers are waiting in line before other servers.

Remark. A real waiting line in which the customers stand one behind the other need not be formed. It suffices that the customers arriving in the system take a number, or that the tasks to be accomplished by the servers be numbered according to their arrival order in the system.

Since the customers arrive one at a time and are served one at a time, the process ${X(t), t \geq 0}$, where $X(t)$ represents the number of customers in the system at time $t$, is a birth and death process. Note that two arbitrary customers cannot leave the system exactly at the same time instant, because the service times are continuous random variables. The balance equations of the system are the following (see Fig. $6.2$ for the $M / M / 2$ model):
$\underline{\text { state } j}$ departure rate from $j=$ arrival rate to $j$
\begin{aligned} 0 & \lambda \pi_0 &=\mu \pi_1 \ 0<k<s &(\lambda+k \mu) \pi_k &=(k+1) \mu \pi_{k+1}+\lambda \pi_{k-1} \ k \geq s &(\lambda+s \mu) \pi_k &=s \mu \pi_{k+1}+\lambda \pi_{k-1} \end{aligned}

## 数学代写|随机过程Stochastic Porcesses代考|The model M/M/s/c and loss systems

As we mentioned in Subsection $6.2 .2$, in reality the capacity $c$ of a waiting system is generally finite. In the case of the $M / M / s / c$ model, we can use the results on birth and death processes to calculate the limiting probabilities of the process.

Example 6.3.2. Consider the queueing system $M / M / 2 / 3$ for which $\lambda=2$ and $\mu=4$. The balance equations of the system are
$$\begin{array}{cr} \text { state } j & \text { departure rate from } j= \ 0 & 2 \pi_0 \stackrel{(0)}{=} 4 \pi_1 \ 1 & (2+4) \pi_1 \stackrel{(1)}{=} 2 \pi_0+(2 \times 4) \pi_2 \ 2 & (2+2 \times 4) \pi_2 \stackrel{(2)}{=} 2 \pi_1+(2 \times 4) \pi_3 \ 3 & (2 \times 4) \pi_3 \stackrel{(3)}{=} 2 \pi_2 \end{array}$$
We can directly solve this system of linear equations. Equation $(0)$ yields $\pi_1=\frac{1}{2} \pi_0$. Substituting into (1), we obtain $\pi_2=\frac{1}{4} \pi_1=\frac{1}{8} \pi_0$. Next, we deduce from Eq. (3) that $\pi_3=\frac{1}{4} \pi_2=\frac{1}{32} \pi_0$. We can then write that

$$\pi_0+\frac{1}{2} \pi_0+\frac{1}{8} \pi_0+\frac{1}{32} \pi_0=1 \quad \Longrightarrow \quad \pi_0=\frac{32}{53}, \pi_1=\frac{16}{53}, \pi_2=\frac{4}{53}, \pi_3=\frac{1}{53}$$
Note that this solution also satisfies Eq. (2).
The average number of customers in the system in equilibrium is given by
$$\bar{N}=\sum_{k=0}^3 k \pi_k=\frac{16}{53}+2 \times \frac{4}{53}+3 \times \frac{1}{53}=\frac{27}{53}$$
from which the average time that an entering customer spends in the system is
$$\bar{T}=\frac{\bar{N}}{\lambda_e}=\frac{27 / 53}{2\left(1-\pi_3\right)}=\frac{27 / 53}{2(52 / 53)}=\frac{27}{104} \simeq 0.2596$$
We also find, in particular, that the limiting probability that the system is not empty is given by $1-\pi_0=\frac{21}{53}$.

## 数学代写|随机过程Stochastic Porcesses代考|The model M/M/s

state $j$ 出发率从 $j=$ 到达率 $j$
$0 \lambda \pi_0 \quad=\mu \pi_1 0<k<s(\lambda+k \mu) \pi_k \quad=(k+1) \mu \pi_{k+1}+\lambda \pi_{k-1} k \geq s(\lambda+s \mu) \pi_k \quad=s \mu \pi_{k+1}+\lambda \pi_{k-1}$

## 数学代可|随机过程Stochastic Porcesses代考|The model $M / M / s / c$ and loss systems

state $j \quad$ departure rate from $j=0 \quad 2 \pi_0 \stackrel{(0)}{=} 4 \pi_1 1 \quad(2+4) \pi_1 \stackrel{(1)}{=} 2 \pi_0+(2 \times 4) \pi_2 2 \quad(2+2 \times 4) \pi_2 \stackrel{(2)}{=} 2 \pi_1+(2 \times 4) \pi_3 3 \quad(2 \times 4) \pi_3 \stackrel{(3)}{=} 2 \pi_2$

$$\pi_0+\frac{1}{2} \pi_0+\frac{1}{8} \pi_0+\frac{1}{32} \pi_0=1 \quad \Longrightarrow \quad \pi_0=\frac{32}{53}, \pi_1=\frac{16}{53}, \pi_2=\frac{4}{53}, \pi_3=\frac{1}{53}$$

$$\bar{N}=\sum_{k=0}^3 k \pi_k=\frac{16}{53}+2 \times \frac{4}{53}+3 \times \frac{1}{53}=\frac{27}{53}$$

$$\bar{T}=\frac{\bar{N}}{\lambda_e}=\frac{27 / 53}{2\left(1-\pi_3\right)}=\frac{27 / 53}{2(52 / 53)}=\frac{27}{104} \simeq 0.2596$$

## MATLAB代写

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